Answer
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Hint: Start by taking the selling price as x for both the conditions stated in the question. Try to form the relation between the two selling prices as stated in the question and apply the loss(%) formula in order to get one of the unknown values and use the relation to find the other one . Apply the profit(%) formula to get the final desired value.
Complete step-by-step answer:
Given, loss of 40% at selling price$\dfrac{2}{5}th$ of earlies selling price.
Let , the first selling price be =${x_1}$
Then , second selling price be =${x_2}$
$\therefore {x_2} = \dfrac{2}{5}{x_1} \to (1)$
We know , ${\text{Loss}}(\% ) = \dfrac{{{\text{Cost price - selling price}}}}{{{\text{Cost price}}}} \times 100$
$
40 = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow \dfrac{{40}}{{100}} = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow 0.4 = 1 - \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 1 - 0.4 \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 0.6 \\
\Rightarrow {x_2} = 0.6 \times {\text{C}}{\text{.P}}{\text{.}} \\
$
Now , substituting this value in relation 1, we get
$
{x_2} = \dfrac{2}{5}{x_1} \\
0.6 \times {\text{C}}{\text{.P}}{\text{. = 0}}{\text{.4}} \times {x_1} \\
\Rightarrow {x_1} = 1.5 \times {\text{C}}{\text{.P}}{\text{.}} \\
\\
$
Now, we know ${\text{Profit}}(\% ) = \dfrac{{{\text{selling price - Cost price}}}}{{{\text{Cost price}}}} \times 100$
Profit earned while selling at first selling price,
$
{\text{Profit}}(\% ) = \dfrac{{{\text{1}}{\text{.5}} \times {\text{C}}{\text{.P}}{\text{. - C}}{\text{.P}}{\text{.}}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (1.5 - 1) \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (0.5) \times 100 \\
\therefore {\text{Profit}}(\% ) = 50\% \\
$
Therefore , the profit earned is 50% when sold at an earlier price.
Note: All the formulas related to profit and loss must be well known to the students . Attention must be given while substituting the values in the equation and also while forming the required relations.
Always remember if you get profit as negative value , then it is a loss and not a profit.
Complete step-by-step answer:
Given, loss of 40% at selling price$\dfrac{2}{5}th$ of earlies selling price.
Let , the first selling price be =${x_1}$
Then , second selling price be =${x_2}$
$\therefore {x_2} = \dfrac{2}{5}{x_1} \to (1)$
We know , ${\text{Loss}}(\% ) = \dfrac{{{\text{Cost price - selling price}}}}{{{\text{Cost price}}}} \times 100$
$
40 = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow \dfrac{{40}}{{100}} = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow 0.4 = 1 - \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 1 - 0.4 \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 0.6 \\
\Rightarrow {x_2} = 0.6 \times {\text{C}}{\text{.P}}{\text{.}} \\
$
Now , substituting this value in relation 1, we get
$
{x_2} = \dfrac{2}{5}{x_1} \\
0.6 \times {\text{C}}{\text{.P}}{\text{. = 0}}{\text{.4}} \times {x_1} \\
\Rightarrow {x_1} = 1.5 \times {\text{C}}{\text{.P}}{\text{.}} \\
\\
$
Now, we know ${\text{Profit}}(\% ) = \dfrac{{{\text{selling price - Cost price}}}}{{{\text{Cost price}}}} \times 100$
Profit earned while selling at first selling price,
$
{\text{Profit}}(\% ) = \dfrac{{{\text{1}}{\text{.5}} \times {\text{C}}{\text{.P}}{\text{. - C}}{\text{.P}}{\text{.}}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (1.5 - 1) \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (0.5) \times 100 \\
\therefore {\text{Profit}}(\% ) = 50\% \\
$
Therefore , the profit earned is 50% when sold at an earlier price.
Note: All the formulas related to profit and loss must be well known to the students . Attention must be given while substituting the values in the equation and also while forming the required relations.
Always remember if you get profit as negative value , then it is a loss and not a profit.
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